How many pounds?

Question

I gave each of my four children a number of pound coins and that was all money I had. To the first I gave some, to the second the third of what remained, to the third 5/8 (five eighth) of what remained and to the fourth the balance which equals 2/5 (two fifths) of the first child share. The total money I had was less than 50 pounds. How many pounds did I have and how much did each child get?

Answer 1

the total money is 39 pounds.
first child = 15 pounds
second child = 8 pounds
third child = 10 pounds
fourth child = 6 pounds

Answer 2

Let x be the total money.
Let y be the money the first child received.
Then,
(x - y) / 3 for the second child
5(x - y) / 12 for the third child
(x - y) / 4 for the fourth child

2y / 5 = (x - y) / 4
5x = 13y

Then,
5x / 13 for the first child
8x / 39 for the second child
10x / 39 for the third child
2x / 13 for the fourth child

x must be an integer (you cannot divide a coin and give it!!!)
then x must be divisible by 39.
The minimum value which is divisible by 39 is itself!

Therefore, each received 15, 8, 10, and 6 pounds respectively.

Answer 3

The previous answers are good, except for the mathematically unsound amount of guessing...
Here's the maths:
A system of 4 equalities and one inequality of five unknowns:

T=a+b+c+d
b=1/3(T-a)
c=5/8(T-a-b)
d=2/5(a)
T<50

The equations resolve to:

a=3/15, b=8/39, c=10/39, d=2/13. Taking the inequality as an additional assumption, the total amount T has to be 39, since a multiple of 39 would be superior to 50, an a fraction would take b and c out the integer domain, unusual for a number of pound coins.

Answer 4

Think of the last child - they get 3/8 times 2/3 of what's left after the first portion has been taken away - that's 6/24. If we guess at six pounds for them the others work out in proportion.
First person gets 15 coins, 2nd gets 8 coins, 3rd gets 10. last one gets 6 39 pounds in total

Answer 5

You had 47. First child 5, second 12, third 5, fourth 1

Answer 6

Total 39
1st 15
2nd 8
3rd 10
4th 6

Answer 7

let A= total amount of cash
x = amount first child gets.
Then C1 receives x
C2 receives (A-x)/3
C3 receives (A-x-(A-x)/3) * 5/8
C4 receives 2/5 * x

So. A = x + (A-x)/3 + (A-x-(A-x)/3) * 5/8 + 2/5 * x
A = 0.65x + 0.75A
0.25A=0.65x
A=2.6x

Now... we just need to find out which values of x give whole numbers for A and for the amounts each child recieves. x= 15 works.

So you had 2.6*15=£39 to start with,
C1 receives £15
C2 receives £8
C3 receives £10
C4 receives £6

Answer 8

a=(5/13)N=first child's money (c1)

N<50 => N=39 as N=13 or N=26 (multiples of 13) doesn't give out integers.

c1 gets 15
c2 8
c3 10
c4 6

nice

Answer 9

What did the first child do to deserve so much more than the others?!

Answer 10

1. I don't really care.
2. Do your homework yourself you lazy person. It's not a difficult question if you can be bothered to set up the algebraic equations.